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Concepts Of Physics MCQ Edition [Volume 1]PhysicsCentre of Mass, Linear momentum, Collision

A block of mass m is positioned on a triangular block of mass M , which in turn rests on a horizontal surface as shown in the figure. Assuming all surfaces are frictionless, determine the velocity of the triangular block when the smaller block reaches the bottom end.

Options

  1. A[ 2M^2 gh ^2 (M + m)(m + M ^2 ) ]^ 1/2
  2. B[ 2m^2 gh ^2 (M + m)(M + m ^2 ) ]^ 1/2
  3. C[ 2m^2 gh ^2 (M + m)(M - m ^2 ) ]^ 1/2
  4. D[ 2m^2 gh ^2 (M + m)(M + m ^2 ) ]^ 1/2

Correct answer

D. [ 2m^2 gh ^2 (M + m)(M + m ^2 ) ]^ 1/2

Step-by-step solution

Let V be the velocity of the triangular block of mass M towards the left. Let v_r be the velocity of the block of mass m relative to the triangular block, directed down the incline. The velocity of the block m relative to the ground is v = v _r + V . Taking the right direction as positive x and upward as positive y , the velocity vectors are: V = -V i v _r = v_r i - v_r j v = (v_r - V) i - v_r j Since there is no external force in the horizontal direction, the horizontal momentum of the system is conserved. Initial

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